drag each triangle to the correct category.

for each set of side lengths, classify the triangle based on whether it is a right triangle.

right triangle | not a right triangle

$8, 7, \sqrt{113}$
$4, 4, \sqrt{32}$
$7, 8, 15$
$4, 4, \sqrt{24}$
$4, 10, 11$

Question

drag each triangle to the correct category.

for each set of side lengths, classify the triangle based on whether it is a right triangle.

right triangle | not a right triangle

$8, 7, \sqrt{113}$
$4, 4, \sqrt{32}$
$7, 8, 15$
$4, 4, \sqrt{24}$
$4, 10, 11$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [], "new_concepts": [ "Pythagorean Theorem Converse", "Triangle Classification" ], "current_concepts": [ "Pythagorean Theorem Converse", "Triangle Classification" ] } ## Identify the longest side for each set $$ \begin{aligned} &\text{Set 1: } 8, 7, \sqrt{113} \implies 8^2 = 64, 7^2 = 49, (\sqrt{113})^2 = 113. \text{ Since } 113 > 64 \text{ and } 113 > 49, \text{ the longest side is } \sqrt{113}.\ &\text{Set 2: } 4, 4, \sqrt{32} \implies 4^2 = 16, 4^2 = 16, (\sqrt{32})^2 = 32. \text{ Since } 32 > 16, \text{ the longest side is } \sqrt{32}.\ &\text{Set 3: } 7, 8, 15 \implies \text{Note: } 7 + 8 = 15. \text{ This does not form a triangle by the triangle inequality } a + b > c.\ &\text{Set 4: } 4, 4, \sqrt{24} \implies 4^2 = 16, 4^2 = 16, (\sqrt{24})^2 = 24. \text{ Since } 24 > 16, \text{ the longest side is } \sqrt{24}.\ &\text{Set 5: } 4, 10, 11 \implies \text{The longest side is } 11. \end{aligned} $$ ## Apply the Pythagorean converse $a^2 + b^2 = c^2$ $$ \begin{aligned} &\text{Set 1: } 7^2 + 8^2 = 49 + 64 = 113 = (\sqrt{113})^2 \implies \text{Right Triangle}\ &\text{Set 2: } 4^2 + 4^2 = 16 + 16 = 32 = (\sqrt{32})^2 \implies \text{Right Triangle}\ &\text{Set 3: } 7 + 8 = 15 \implies \text{Not a Triangle (Not a Right Triangle)}\ &\text{Set 4: } 4^2 + 4^2 = 16 + 16 = 32 e 24 \implies \text{Not a Right Triangle}\ &\text{Set 5: } 4^2 + 10^2 = 16 + 100 = 116 e 11^2 \implies \text{Not a Right Triangle} \end{aligned} $$ ## Categorize the sets $$ \begin{aligned} &\text{Right Triangle: } \{8, 7, \sqrt{113}\}, \{4, 4, \sqrt{32}\}\ &\text{Not a Right Triangle: } \{7, 8, 15\}, \{4, 4, \sqrt{24}\}, \{4, 10, 11\} \end{aligned} $$ | Right Triangle | Not a Right Triangle | |---|---| | $8, 7, \sqrt{113}$ | $7, 8, 15$ | | $4, 4, \sqrt{32}$ | $4, 4, \sqrt{24}$ | | | $4, 10, 11$ | { "subject": "Mathematics", "question_type": "Free Response", "knowledge_point": [ "Mathematics", "Geometry", "Pythagorean Theorem Converse" ] }

Right Triangle	Not a Right Triangle
\(4, 4, \sqrt{32}\)	\(4, 4, \sqrt{24}\)
	\(4, 10, 11\)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Identify the longest side for each set

Apply the Pythagorean converse \(a^2 + b^2 = c^2\)

Categorize the sets

Answer: